One attractive feature of a differential amplifier is its ability to reject signals that arc common to both inputs. Since the outputs are amplified versions of the difference between the inputs, any voltage component that appears identically in both signal inputs will be “differenced out,” that is, will have zero level in the outputs. (We have already seen that the outputs are exactly 0 when both inputs arc identical. inphase signals.) Any de or ac voltage that appears simultaneously in both signal inputs is called a common-mode signal. The ability of an amplifier to suppress. or zero-out. common-mode signals is called common-mode rejection An example of a common-mode signal whose rejection is desirable is electrical noise induced in both signal lines. a frequent occurrence when the lines arc routed together over long paths. Another example is a de level common to both inputs, or common de fluctuations caused by power-supply variations.

In the ideal differential amplifier, any common-mode signal will be completely cancelled out and therefore have no effect on the output signals. In practical amplifiers, which we will discuss in the next section, mismatched components and certain other non ideal conditions result in imperfect cancellation of common-mode signals. Figure 12-15 shows a differential amplifier in which a common-mode signal is applied to both inputs. Ideally, the output voltages should be 0, but in fact some small component of Un. may appear. The differential code gain, Am” is defined to be the ratio of the output difference voltage caused by the commonmode signal to the common-mode signal itself.

We can also define a single-ended common-mode gain as the ratio of (Vpl)”” or (V 2),/II to V, Obviously the ideal amplifier has common-mode gain equal to O. A widely used specification and figure of merit for a differential amplifier is its common-mode rejection ratio (CMRR), defined to be the ratio of the magnitude of its differential (difference-mode) gain Ad to the magnitude of its common-mode gain:

When the inputs to a certain differential amplifier arc Vii = 0.1 sin cui and Vi2 = -0.1 sin to]; it is found that the outputs are v.” == -5 sin wi and V,,! ::: 5 sin wf. When both inputs arc 2 sin wi, the outputs arc VI’I= -0.05 sin tot and = 0.05 sin WI. Find the CMRR in dB.

Solution“, We will use the peak values of the various signals for our gain computations, but note carefully how the minus signs arc used to preserve phase relations. The diffrance-mode gain


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